## College Algebra (6th Edition)

$$x\approx-1.14$$
$$e^{1-5x}=793$$ Here we would take the logarithm of both sides. Since the left side involves $e$, we would take the natural logarithm. $$\ln(e^{1-5x})=\ln793$$ $$(1-5x)\ln e=\ln793$$ $$1-5x=\ln793$$ (for $\ln e=1$) $$5x=1-\ln793$$ $$x=\frac{1-\ln793}{5}$$ $$x\approx-1.14$$